Adding an extra temperature sensor to the Asus A7V  Computer Motherboard.

How to substitute one thermister for another with a different (higher) beta.

(And very possibly other boards that use thermistors)

I recently upgraded my Computer which uses a A7V V1.04 motherboard by installing a 1300 Duron and in the process had to consider the the higher temperatures involved.

I noticed a spare pin header (JTPWR) that the manual refers to as suitable for a power supply temperature sensor. The first thing I did was to try to find the specifications. I had no luck but did find a reference that said that Asus chip involved had a similar specification to the Winborne chip and it's specification showed that a Radio Shack thermistor part number 271-0110 was to be used. If you can get this part you can just wire it up to a suitable connector without any extra resistors so you can  skip this next section..

Data on The 271-0110 was available at http://support.radioshack.com/support_supplies/doc33/33553.pdf

Using available thermistors.
The correct thermistor has a resistance of 10k ohms at 25C , 4.161k at 50C and a "B"value of 3435 +- 1%. The B value in simple terms is how much the resistance changes with temperature. I confirmed that this is the correct part for the A7V by inserting a 4.161k resistor where the thermister goes and getting a reading of 50C.

It was easy to find thermistors with the required 10k ohms at 25C but getting the correct B value was impossible. All of the thermistors that I found had a B that was too high. However by the addition of 2 resistors they could be made to work.

The first step in using an alternate thermistor is to determine its resistance at 50C, usually you get this from a data sheet. The correct part would have a value of 4.161k ohm but so long as yours is lower look on the graph below and determine the value of the "RS" resistor. For example a "Curve Matched" 10k thermistor (Unitherm Type EC95) available from RS Components ( stock # 151-237) for about $2 has a resistance of 3.894k ohm @ 50C and would require an "RS" resistor of .33k ie. 330 ohms. Putting this resistor in series with the thermistor would increase the total resistance and make the readings incorrect but it can be balanced by a resistor that is connected across the thermistor and the RS resistor . This resistor (RP) is calculated as follows:-

RP = 10 + (100/RS) k ohms thus for the above device RP = 10 + (100/.33) k =313.03k ohm

Selecting values from the 1% resistor range you would use RS=330 ohm and RP= 300k ohm. This combination of resistors and a Unitherm EC95 gives a resistance of 4.165k at 50C and 9.986k at 25C and is within .2% of the specification of the desired 271-0110 thermister.

table


circuit
Basic theory.

With the thermistor at 25C  i.e. 10k  there are combinations of RS and RP that maintain the overall resistance of the circuit at 10k. For example with RS=0 (short circuit) RP must be infinite (open circuit) to maintain 10k. In this condition all of any change in the resistance of the thermistor is reflected in the overall resistance of the circuit. With RS=infinite (open circuit) RP must be 10k and in this condition none of any change in the resistance of the thermistor is reflected in the overall resistance of the circuit.  There are intermediate values of RS and RP where the overall resistance of the circuit is still 10k at 25C but less than the full amount of any change in the resistance of the thermistor is reflected in the overall resistance of the circuit. Thus a thermistor with a larger change can be substituted for one with lesser change by adding a couple of resistors.

Putting it together.

If you have the correct thermistor solder it across the wires going to the connector.

Otherwise work out the two resistor values you need and then select the closest values from the 1/4 Watt 1% range.

You will need a connector for the board. I used an old fan connector, it was 3 pin but It fitted on the 2 pin header, I just did not use the extra pin hole. The Power Supply Thermal Sensor Connector (2-pin block marked JTPWR) is located at the top edge of the board near the RAM slots (see manual page 40).

Arrange the two resistors and the thermistor so that they can be soldered and insulated with heat-shrink tubing without shorting.


Resistor RS connected to thermistor.


First lead connected, polarity is not important.


Resistor RP connected from thermistor to red wire and black lead added.

Note that the sleeve should not be able to slide down the thermistor leg. To finish off cover the lot with another heat-shrink sleeve.

To see the temperature using the Mother Board Monitor program select Asus 3 in the temperature menu. Asus Probe software seems not to have the ability to monitor this Asus feature.

You MUST NOT use this temperature probe INSIDE THE POWER SUPPLY as it is DANGEROUS and also electrical noise may be introduced into the computer, also do not use it on the outside of the computer case. I used mine for room temperature by mounting it near a air inlet to the case. This has an advantage over a normal thermometer in that the temperature can be logged to a file by MBM.

When installing the probe make sure that the computer power lead is disconnected from the wall socket and take precautions against static electricity.

Note this method of  substituting one thermistor for another may have general application. You just need a thermistor with the same resistance at 25C and a higher  beta. Unfortunately the mathematics to provide a general formula for any required thermister is beyond me.  The table above was derived from experiment.
Update. A quadratic equation can be obtained by substituting in the formula below and solved using one of the online solvers of quadratic equations.
A = resistance of thermistors at 25C    B = resistance of  thermistor to be emulated at some higher temperature  C = resistance of replacement thermistor (with higher beta)
at the same higher temperature X = series reistor P = parallel resistor

X^2   + (C+A)X + C*A  -   (A-C) *A*B / (A-B)  =  0

Substituting these values from the above example    A = 10  B = 4.161  C = 3.894
gives the following equation
X^2 + 13.8945X - 4.5727 = 0      
which can be solved for example with
http://www.mathsisfun.com/quadratic-equation-solver.html

Quadratic solver

The series resistor is .321k (ignore the negative root)
The parallel resistor P = A + A^2/X  = 320.9k

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